A confidence interval for the mean price shows a $9.30 margin of error. What was the level of confidence? A. 99.9% B. 99.8% C. 99.7% D. 99.6% E. 99.5%
I know that z(a/2) (std dev)/(sqrt n)=9.3
z(a/2)=3.1
And the corresponding z score is 0.9987, but I am struggling because of the (a/2), and I feel like an extra step is required.
The correct answer is Option B: 99.8% Below is the explanation in the image.
Standard deviation $\sigma=\$12$, $n=16$, $E=\$9.30$
\begin{align*} E=Z_{\alpha/2}\cdot\frac\sigma{\sqrt n} &\implies 9.30=Z_{\alpha/2}\cdot\frac{12}{\sqrt{16}} \\ &\implies Z_{\alpha/2}=9.30/3=3.10 \end{align*}
As $$P(2<3.10)=0.9990\implies\alpha/2=1-0.9990=0.001$$ it follows that $\underline{\alpha=0.002}$. So, for the level of confidence $c$ $$c=1-\alpha=1-0.002=0.998\,\therefore\, c=99.8\%$$